3 and 5 Litre Jug Puzzle

Question

You may have seen this one in Die Hard: With a Vengeance... This question is based on the famous 3 and 5 Litre Jug Puzzle, but with a slightly different slant.

Golf up some code that when given an integer between 1 and 100 will provide you with the quickest instructions to measure out into a tank, the corresponding number of litres of water from a fountain, using a 3 litre jug and a 5 litre jug.

There are no gradations on either of the jugs; the fountain is abundant in supply of water, and the tank is assumed to be emptied out upon the start of each execution of the code.

You cannot access water from the tank once it goes into the tank.

The format of execution is as follows:

Input:

4 for example.

Output

Output each numbered step, as shown, followed by a tally of the volumes of the 5L jug, the 3L jug and the tank. Tally format also shown below. The number of steps must also be outputted at the end of the steps.

1) Fill 5L jug

5L: 5, 3L: 0, T: 0

2) Pour from 5L jug into 3L jug

5L: 2, 3L: 3, T: 0

3) Empty 3L jug

5L: 2, 3L: 0, T: 0

4) Pour from 5L jug into 3L jug

5L: 0, 3L: 2, T: 0

5) Fill 5L jug

5L: 5, 3L: 2, T: 0

6) Pour from 5L jug into 3L jug

5L: 4, 3L: 3, T: 0

7) Pour from 5L jug into tank

5L: 0, 3L: 3, T: 4

Volume measured out in 7 turns

Example 2

Input: 8

Output:

1) Fill 5L jug

5L: 5, 3L: 0, T: 0

2) Pour from 5L jug into tank

5L: 0, 3L: 0, T: 5

3) Fill 3L jug

5L: 0, 3L: 3, T: 5

4) Pour from 3L jug into tank

5L: 0, 3L: 0, T: 8

Volume measured out in 4 turns

Conventions

1. Fill xL jug - fills the associated jug to the top from the fountain
2. Empty xL jug - empties the contents of the associated jug into the fountain
3. Pour from xL jug into yL jug - Pours the contents of the xL jug into the yL jug
4. Pour from xL jug into tank - Pours the contents of the xL jug into the tank

Shortest code wins.

Answer 1

Ruby, 407 376 365 331 324 323

This is getting kind of hard to read...

x=y=n=d=0
g=gets.to_i
"#{[43435,102,t=45,t,12,t,12,t,t][g+~d]||12}".chars{|c|n+=1
puts [eval(["x-=t=[3-y,x].min;y+=t"+t=";'Pour from 5L jug into 3L jug'","x=5;'Fill 5L jug'","d+=x;x=0"+t.sub(/3.+/,"tank'")][c.ord%3].tr t='35xy',c<?3?t:'53yx'),"5L: #{x}, 3L: #{y}, T: #{d}"]}while g>d
$><<"Volume measured out in #{n} turns"

Takes input on STDIN. Example run for N=10:

Fill 5L jug
5L: 5, 3L: 0, T: 0
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 5
Fill 5L jug
5L: 5, 3L: 0, T: 5
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 10
Volume measured out in 4 turns

Answer 2

T-SQL 2012: 1410 1302

Another quixotic attempt at a question in SQL, but this one offered an enjoyable opportunity to play with some of the new window function options in version 2012. In addition, it exploits recursive CTEs, which may be nothing impressive in most programming languages, but recursion in SQL is like switching from horse and buggy to a Ferrari.

The engine at the heart of this is in lines 5-12, which uses a recursive CTE and a window function to build a table of most of the numbers needed to solve the problem. Note in particular the test for 3, 4, 6, or 9, which ensures an optimal approach to the solution by 3s from those numbers onward. (Technically, it's a tie for 4 between the 3-1 approach and the 2-2, but doing it this way golfed me a lot of characters.) Then it's a simple matter to join to a lookup table of the optimal steps for different chunks of the problem and use another window function to properly number the steps.

If you don't have MS SQL lying around, play with it on SQLFiddle.

DECLARE @i INT=42,@l VARCHAR(9)='L jug ',@k VARCHAR(99)='into tank
5L: 0, 3L: 0, T: ',@o VARCHAR(99)='
5L: 5, 3L: 0, T: ',@n CHAR(1)='
',@5 VARCHAR(99)=') Pour from 5',@3 VARCHAR(99)=') Pour from 3'
;WITH t AS (SELECT @i i,(@i-@i%5)%5 j
UNION ALL
SELECT i-5,(i-i%5)%5+5 FROM t WHERE i>=5 AND i NOT IN(6,9)
UNION ALL
SELECT i-3,3FROM t WHERE i in(3,4,6,9)
UNION ALL
SELECT i-i,i FROM t WHERE i<3 AND i>0)
SELECT t.i,t.j,v.s,ROW_NUMBER()OVER(PARTITION BY t.j ORDER BY t.i DESC)x,SUM(t.j)OVER(ORDER BY t.i DESC ROWS UNBOUNDED PRECEDING)y INTO #q FROM(VALUES(1,5),(2,3),(3,2),(5,2))v(i,s) JOIN t ON t.j = v.i
SELECT z.b FROM(SELECT ROW_NUMBER()OVER(ORDER BY q.i DESC,w.s)a,CAST(ROW_NUMBER()OVER(ORDER BY q.i DESC,w.s)AS VARCHAR)+w.v+CAST(y-CASE WHEN q.s!=w.s THEN q.j ELSE 0 END AS VARCHAR)b
FROM(VALUES(5,1,') Fill 5'+@l+@o),(5,2,@5+@l+@k),(3,1,') Fill 3'+@l+@n+'5L: 0, 3L: 3, T: '),(3,2,@3+@l+@k),(2,1,') Fill 5'+@l+@o),(2,2,@5+@l+' into 3'+@l+@n+'5L: 2, 3L: 3, T: '),(2,3,@5+@l+@k),(1,1,') Fill 3'+@l+@n+'5L: 0, 3L: 3, T: '),(1,2,@3+@l+'into 5'+@l+@n+'5L: 3, 3L: 0, T: '),(1,3,') Fill 3'+@l+@n+'5L: 3, 3L: 3, T: '),(1,4,@3+@l+'into 5'+@l+@n+'5L: 5, 3L: 1, T: '),(1,5,@3+@l+'into tank'+@o))w(i,s,v)JOIN #q q ON w.i=q.j
UNION
SELECT 99,'Volume measured out in '+CAST(COUNT(*)AS VARCHAR)+' turns'
FROM #q)z

Results for the input 42:

1) Fill 5L jug 
5L: 5, 3L: 0, T: 0
2) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 5
3) Fill 5L jug 
5L: 5, 3L: 0, T: 5
4) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 10 
5) Fill 5L jug 
5L: 5, 3L: 0, T: 10 
6) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 15 
7) Fill 5L jug 
5L: 5, 3L: 0, T: 15 
8) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 20 
9) Fill 5L jug 
5L: 5, 3L: 0, T: 20 
10) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 25 
11) Fill 5L jug 
5L: 5, 3L: 0, T: 25 
12) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 30 
13) Fill 5L jug 
5L: 5, 3L: 0, T: 30 
14) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 35 
15) Fill 5L jug 
5L: 5, 3L: 0, T: 35 
16) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 40 
17) Fill 5L jug 
5L: 5, 3L: 0, T: 40 
18) Pour from 5L jug  into 3L jug 
5L: 2, 3L: 3, T: 40 
19) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 42 
Volume measured out in 9 turns 

Edit:

Golfed out a decent score improvement by

• eliminating an unnecessary +5 in the first row of the CTE, and the WHERE clause it necessitated
• in-lining the VALUES tables, saving costly DECLARE statements
• remembering to convert Windows double-byte CRLFs to Unix style this time.

Answer 3

Javascript: 481

First attempt at golfing, advice appreciated

n=["3L jug","5L jug","tank"];l=[0,0,0];t=[3,5,0];h=0;c=console;function e(d){l[d]=t[d];c.log(++h+") Fill "+n[d]);k()}function m(d,g){s=l[d];f=l[g];b=s+f>t[g];l[g]=b?t[g]:f+s;l[d]=b?s-(t[g]-f):0;c.log(++h+") Pour from "+n[d]+" into "+n[g]);k()}function k(){c.log("5L: "+l[1]+", 3L: "+l[0]+", T: "+l[2])}a=prompt();for(t[2]=a;4<a;)e(1),m(1,2),a-=5;2<a&&(e(0),m(0,2),a-=3);1<a&&(e(1),m(1,0),m(1,2),a=0);0<a&&(e(0),m(0,1),e(0),m(0,1),m(0,2));c.log("Volume measured out in "+h+" turns")

It messes up with some numbers because it doesn't check if it's better to pour 3 or 5, example: 9 gives 9 turns instead of 6, I might fix it later

Paste it in console

From 553 to 481 thanks to @WallyWest

Answer 4

Java, 610

class X{int n,c=0,t=0;public void static main(String[]a){n=Integer.parseInt(a[0]);String s,b,f,k,m,u;b="5L";s="3L";k="tank";u="Fill %s jug\n5L: %d, 3L: %d, T: %d";m="\nPour from %s jug into %s\n5L: %d, 3L: %d, T: %d";f=u+m;for(;n>4;)z(f,2,5,b,5,0,t,b,k,0,0,t+=5);while(n!=0){if(n==1)z(f+f+m,5,1,s,0,3,t,s,b,3,0,t,s,3,3,t,s,b,5,1,t,s,k,5,0,t+1);if(n==3)z(f,2,3,s,0,3,t,s,k,0,0,t+3);z(f+m,3,2,b,5,0,t,b,s,2,3,t,b,k,0,3,t+=2);if(n==2)z("Empty 3L jug\n5L: 0, 3L: 0,T: %d",1,0,t)}z("Volume measured out in %d turns",0,0,c)}void z(String s,int o,int w,Object...a){c+=o;n-=w;System.out.println(String.format(s,a))}}

I took the solution of Sumedh and golfed it. I wanted to put it in the comments but my reputation isn't enough :( . It's a 40% less, I think it should at least be shared. Still far from first though...

Here is ungolfed:

    class X{
    int n,c=0,t=0;
    public void static main(String[] a){
        n=Integer.parseInt(a[0]);
        String s,b,f,k,m,u;
        b="5L";
        s="3L";
        k="tank";
        u="Fill %s jug\n5L: %d, 3L: %d, T: %d";
        m="\nPour from %s jug into %s\n5L: %d, 3L: %d, T: %d";
        f=u+m;
        for(;n>4;)z(f,2,5,b,5,0,t,b,k,0,0,t+=5);
        while(n!=0)
        {
            if(n==1)z(f+f+m,5,1,s,0,3,t,s,b,3,0,t,s,3,3,t,s,b,5,1,t,s,k,5,0,t+1);
            if(n==3)z(f,2,3,s,0,3,t,s,k,0,0,t+3); 
            z(f+m,3,2,b,5,0,t,b,s,2,3,t,b,k,0,3,t+=2);
            if(n==2)z("Empty 3L jug\n5L: 0, 3L: 0,T: %d",1,0,t);
        }
        z("Volume measured out in %d turns",0,0,c);
    }
    void z(String s,int o, int w,Object... a){
        c+=o;
        n-=w;
        System.out.println(String.format(s,a));
    }
}

NB: it works only on the first run. Rerun it and the result will be wrong (due to a global variable).

The following version is safe, but we lose 2 char, going from 610 to 612:

    class X{
    int n,c,t;
    public void static main(String[] a){
        n=Integer.parseInt(a[0]);
        String s,b,f,k,m,u;
        t=c=0;
        b="5L";
        s="3L";
        k="tank";
        u="Fill %s jug\n5L: %d, 3L: %d, T: %d";
        m="\nPour from %s jug into %s\n5L: %d, 3L: %d, T: %d";
        f=u+m;
        for(;n>4;)z(f,2,5,b,5,0,t,b,k,0,0,t+=5);
        while(n!=0)
        {
            if(n==1)z(f+f+m,5,1,s,0,3,t,s,b,3,0,t,s,3,3,t,s,b,5,1,t,s,k,5,0,t+1);
            if(n==3)z(f,2,3,s,0,3,t,s,k,0,0,t+3); 
            z(f+m,3,2,b,5,0,t,b,s,2,3,t,b,k,0,3,t+=2);
            if(n==2)z("Empty 3L jug\n5L: 0, 3L: 0,T: %d",1,0,t);
        }
        z("Volume measured out in %d turns",0,0,c);
    }
    void z(String s,int o, int w,Object... a){
        c+=o;
        n-=w;
        System.out.println(String.format(s,a));
    }
}

Sample output for N=69:

Fill 5L jug
5L: 5, 3L: 0, T: 0
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 5
Fill 5L jug
5L: 5, 3L: 0, T: 5
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 10
Fill 5L jug
5L: 5, 3L: 0, T: 10
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 15
Fill 5L jug
5L: 5, 3L: 0, T: 15
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 20
Fill 5L jug
5L: 5, 3L: 0, T: 20
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 25
Fill 5L jug
5L: 5, 3L: 0, T: 25
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 30
Fill 5L jug
5L: 5, 3L: 0, T: 30
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 35
Fill 5L jug
5L: 5, 3L: 0, T: 35
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 40
Fill 5L jug
5L: 5, 3L: 0, T: 40
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 45
Fill 5L jug
5L: 5, 3L: 0, T: 45
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 50
Fill 5L jug
5L: 5, 3L: 0, T: 50
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 55
Fill 5L jug
5L: 5, 3L: 0, T: 55
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 60
Fill 5L jug
5L: 5, 3L: 0, T: 60
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 65
Fill 5L jug
5L: 5, 3L: 0, T: 65
Pour from 5L jug into 3L
5L: 2, 3L: 3, T: 65
Pour from 5L jug into tank
5L: 0, 3L: 3, T: 67
Empty 3L jug
5L: 0, 3L: 0,T: 67
Fill 5L jug
5L: 5, 3L: 0, T: 67
Pour from 5L jug into 3L
5L: 2, 3L: 3, T: 67
Pour from 5L jug into tank
5L: 0, 3L: 3, T: 69
Volume measured out in 33 turns

Answer 5

Java: 984

Here's the code

class X{public static void main(String[] s){int n=Integer.parseInt(s[0]);int t=0;int c=0;while(n>4){n-=5;System.out.println("Fill 5L jug\n5L: 5, 3L: 0, T: "+t+"\nPour from 5L jug into tank\n5L: 0, 3L: 0, T: "+(t+5));t+=5;c+=2;}while(n!=0){switch(n){case 1:System.out.println("Fill 3L jug\n5L: 0, 3L: 3, T: "+t+"\nPour from 3L jug into 5L jug\n5L: 3, 3L: 0, T: "+t+"\nFill 3L jug\n5L: 3, 3L: 3, T: "+t+"\nPour from 3L jug into 5L jug\n5L: 5, 3L: 1, T: "+t+"\nPour from 3L jug into tank\n5L: 5, 3L: 0, T: "+(t+1));n=0;c+=5;break;case 3:System.out.println("Fill 3L jug\n5L: 0, 3L: 3, T: "+t+"\nPour from 3L jug into tank\n5L: 0, 3L: 0, T: "+(t+3));n=0;c+=2;break;default:System.out.println("Fill 5L jug\n5L: 5, 3L: 0, T: "+t+"\nPour from 5L jug into 3L jug\n5L: 2, 3L: 3, T: "+t+"\nPour from 5L jug into tank\n5L: 0, 3L: 0, T: "+(t+2));n-=2;c+=3;t+=2;if(n==2){System.out.println("Empty 3L jug\n5L: 0, 3L: 0,T: "+t);c++;}break;}}System.out.println("Volume measured out in "+c+" turns");}}

Input is from command line. for example: java X 4

Answer 6

Python 2.7 - 437

Not the shortest code, but I think this is the most optimal way of solving this.

As I stated in the comments, the most optimal way to calculate this:

1. Take as many chunks of 5L as possible - divmod(amount,5). This will give you one of 4,3,2,1 as the remainder.
2. Take 3 (if possible) from the remainder.

3. Which leaves either 1 or 2 as the remainder. Use the optimal solution for either which can be known ahead of time as:

   1. 1L, 5 steps : 3L -> 5L, 3L -> 5L, leaving 1L in the 3L, 3L (holding 1L) -> tank
   2. 2L, 3 steps : 5L -> 3L, leaves 2L in the 5L, 5L (holding 2L) -> tank

The code:

j,T="%dL jug","tank"
A="\n5L: %d, 3L: %d, T: %d"
F,P="Fill "+j+A,"Pour from "+j+" into %s"+A
f,r=divmod(input(),5)
o,t=f*5,[]
for i in range(f):o+=[F%(5,5,0,5*i),P%(5,T,0,0,5*i+5)]
if r>2:o+=[F%(3,0,3,t),P%(3,T,0,0,t+3)];r-=3;t+=3
if r==2:o+=[F%(5,5,0,t),P%(5,j%3,2,3,t),P%(5,T,0,3,t+2)]
if r==1:o+=[F%(3,0,3,t),P%(3,j%5,3,0,t),F%(3,3,3,t),P%(3,j%5,5,1,t),P%(3,T,5,0,t+1)]
print"\n".join(o),'\n',"Volume measured out in %d turns"%len(o)

And an output for 4L in 7 steps:

Fill 3L jug
5L: 0, 3L: 3, T: 0
Pour from 3L jug into tank
5L: 0, 3L: 0, T: 3
Fill 3L jug
5L: 0, 3L: 3, T: 3
Pour from 3L jug into 5L jug
5L: 3, 3L: 0, T: 3
Fill 3L jug
5L: 3, 3L: 3, T: 3
Pour from 3L jug into 5L jug
5L: 5, 3L: 1, T: 3
Pour from 3L jug into tank
5L: 5, 3L: 0, T: 4
Volume measured out in 7 turns

Answer 7

Smalltalk (Smalltalk/X), 568 560 516

input in n:

    T:=j:=J:=c:=0.m:={'Pour from'.' into'.' 3L jug'.' 5L jug'.[j:=j+3.'Fill'].[J:=J+5.'Fill'].[t:=j.j:=0.''].[t:=J.J:=0.''].[r:=j min:5-J.j:=j-r.J:=J+r.''].[r:=J min:3-j.J:=J-r.j:=j+r.''].[T:=T+t.' into tank'].[c:=c+1.'\5L: %1 3L: %2 T: %3\'bindWith:J with:j with:T].['Volume measured out in %1 turns'bindWith:c]}.[n>=0]whileTrue:[s:=n.n:=0.(s caseOf:{0->[n:=-1.'<'].1->'42;02813;42;02813;062:;'.2->'53;03912;073:;'.3->'42;062:;'.4->[n:=1.'42;062:;']}otherwise:[n:=s-5.'53;073:;'])do:[:c|(m at:c-$/)value withCRs print]]

boy this is definitely the most obfuscated program I've ever written...

Edit: Some other Smalltalks may not allow autodeclared workspace variables and you'll have to prepend declarations. Also bindWith: may be different (expandWith:'<p>').

sample output for n=17:

Fill 5L jug 
5L: 5, 3L: 0, T: 0
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 5
Fill 5L jug 
5L: 5, 3L: 0, T: 5
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 10
Fill 5L jug 
5L: 5, 3L: 0, T: 10
Pour from 5L jug into tank
5L: 0, 3L: 0, T: 15
Fill 5L jug 
5L: 5, 3L: 0, T: 15
Pour from 5L jug into 3L jug 
5L: 2, 3L: 3, T: 15
Pour from 5L jug into tank
5L: 0, 3L: 3, T: 17
Volume measured out in 9 turns

Answer 8

C, 567 609

#define r printf
#define l r("5L: %d, 3L: %d, T: %d\n", a, b, T);
#define j(x,y,z,w) r("%d) "#x" %dL jug\n", i++, y),z=w,l
#define e j(Empty,3,b,0)
#define f j(Fill,5,a,5)
#define g j(Fill,3,b,3)
#define o(x,y,z,w) r("%d) Pour from %dL jug into "x"\n", i++, y,z),w;l
#define t(x,y) T+=x,o("tank",y,0,x=0)
#define p(x) o("%dL jug",5,3,(a-=x,b+=x))
int N,T,i;q(x){int a=0,b=0;switch(x){case 5:f t(a,5) break;case 3:g t(b,3) break;case 1:case 2:case 4:f if(x-2){e p(2)f p(1)if(x-4){e p(3)}}t(a,5)}N-=x;}main(){T=0,i=1,scanf("%d",&N);while(N>5)q((N-6)&&(N-9)?5:3);q(N);r("Volume measured out in %d turns",i-1);}

previous invalid version:

#define r printf
#define l r("5L: %d, 3L: %d, T: %d\n", a, b, T);
#define j(x,y,z,w) r("%d) "#x" %dL jug\n", i++, y),z=w,l
#define e j(Empty,3,b,0)
#define f j(Fill,5,a,5)
#define g j(Fill,3,b,3)
#define o(x,y,z,w) r("%d) Pour from %dL jug into "x"\n", i++, y,z),w;l
#define t o("tank",5,0,a=0)
#define p(x) o("%dL jug",5,3,(a-=x,b+=x))
int N,T,i;q(x){int a=0,b=0;switch(x){case 5:f t break;case 3:g t break;case 1:case 2:case 4:f if(x-2){e p(2)f p(1)if(x-4){e p(3)}}t}N-=x;}main(){T=0,i=1,scanf("%d",&N);while(N>5)q(5);q(N);r("Volume measured out in %d turns",i-1);}

and here is the code degolfed:

#define r printf
#define l r("5L: %d, 3L: %d, T: %d\n", a, b, T);
#define j(x,y,z,w) r("%d) "#x" %dL jug\n", i++, y),z=w,l
#define e j(Empty,3,b,0)
#define f j(Fill,5,a,5)
#define g j(Fill,3,b,3)
#define o(x,y,z,w) r("%d) Pour from %dL jug into "x"\n", i++, y,z),w;l
#define t o("tank",5,0,a=0)
#define p(x) o("%dL jug",5,3,(a-=x,b+=x))
int N,T,i;
q(x)
{
    int a=0,b=0;
    switch(x)
    {
        case 5:
            f
            t 
            break;
        case 3:
            g
            t
            break;
        case 1:
        case 2:
        case 4:
            f
            if(x-2)
            {
                e
                p(2)
                f
                p(1)
                if(x-4)
                {
                    e
                    p(3)
                }
            }
            t
    }
    N-=x;
}
main()
{
    T=0,i=1,scanf("%d",&N);
    while(N>
    5)q(5);
    q(N);
    r("Volume measured out in %d turns",i-1);
}

Answer 9

C (480 465 bytes)

#define P printf(
#define O(x) P"%d) Pour from %dL jug into "x"\n"
#define S P"5L: %d, 3L: %d, T: %d\n",F,H,g);}
F,H,s,g,x;l(j){P"%d) Fill %dL jug\n",++s,j);St(j,o,m){O("%dL jug"),++s,j,(j^5)?5:3);Se(j,i){O("tank"),++s,j);Smain(){scanf("%d",&x);while(x>4){x-=5;l(F=5);g+=5;e(5,F=0);}while(x>2){x-=3;l(H=3);g+=3;e(3,H=0);}(x^2)?(x^1)?0:(l(H=3),t(3,H=0,F=3),l(H=3),t(3,H=1,F=5),g++,e(3,H=0)):(l(F=5),t(5,F=2,H=3),g+=2,e(5,F=0));P"Volume measured out in %d turns",s);}

Optimal version (adds 10 bytes)

#define P printf(
#define O(x) P"%d) Pour from %dL jug into "x"\n"
#define S P"5L: %d, 3L: %d, T: %d\n",F,H,g);}
F,H,s,g,x;l(j){P"%d) Fill %dL jug\n",++s,j);St(j,o,m){O("%dL jug"),++s,j,(j^5)?5:3);Se(j,i){O("tank"),++s,j);Smain(){scanf("%d",&x);while(x>4&&x^6&&x^9){x-=5;l(F=5);g+=5;e(5,F=0);}while(x>2){x-=3;l(H=3);g+=3;e(3,H=0);}(x^2)?(x^1)?0:(l(H=3),t(3,H=0,F=3),l(H=3),t(3,H=1,F=5),g++,e(3,H=0)):(l(F=5),t(5,F=2,H=3),g+=2,e(5,F=0));P"Volume measured out in %d turns",s);}

Likely more golfing to be done here - the output functions were killing me. This should give the optimal solution (least number of steps). Similar to other code here, it fills and empties 5L jugs until it gets below 5 and then switches to 3L jugs. It tests for 2 special cases (6 and 9) and if it finds them switches to 3L jugs. The instructions for getting 1L and 2L are hard coded.

More readable version:

#define P printf(
#define O(x) P"%d) Pour from %dL jug into "x"\n"
#define S P"5L: %d, 3L: %d, T: %d\n",F,H,g);}
F,H,s,g,x;
l(j)
{
    P"%d) Fill %dL jug\n",++s,j);S

t(j,o,m)
{
    O("%dL jug"),++s,j,(j^5)?5:3);S

e(j,i)
{
    O("tank"),++s,j);S

main()
{
    scanf("%d",&x);
    //while(x>4&&x^6&&x^9)     <--optimal version
    while(x>4)
    {
        x-=5;l(F=5);g+=5;e(5,F=0);
    }
    while(x>2)
    {
        x-=3;l(H=3);g+=3;e(3,H=0);
    }
    (x^2)?
        (x^1)?  
            0
             :
            (l(H=3),t(3,H=0,F=3),l(H=3),t(3,H=1,F=5),g++,e(3,H=0))
             :(l(F=5),t(5,F=2,H=3),g+=2,e(5,F=0));
    P"Volume measured out in %d turns",s);
}

Edits:

• Removed 10 bytes that gave the optimum performance for the scored version based on the OP's clarification.
• Shave 5 bytes by converting function to definition.

Test output for n = 11 (optimal version):

11
1) Fill 5L jug
5L: 5, 3L: 0, T: 0
2) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 5
3) Fill 3L jug
5L: 0, 3L: 3, T: 5
4) Pour from 3L jug into tank
5L: 0, 3L: 0, T: 8
5) Fill 3L jug
5L: 0, 3L: 3, T: 8
6) Pour from 3L jug into tank
5L: 0, 3L: 0, T: 11
Volume measured out in 6 turns

Answer 10

T-SQL(2012): 794 689 580

Inspired by @Jonathan-Van-Matre 's T-SQL answer in combination with @Lego-Stormtroopr's algorithm. I wanted to do this because I enjoyed the 99 Bottles of Beer challenge so much.

I tried to keep window (OVER) functions at a minimum in preference of math/bool functions.

SQLFiddle is here.

WITH n AS(SELECT 11 n UNION ALL SELECT n-IIF(n>4,5,3)FROM n WHERE n>2)SELECT n, a,LEN(a)L,i=IDENTITY(INT,1,1),'L jug'j INTO #t FROM n JOIN(VALUES(3303),(33900),(5550),(55900),(2550),(259323),(25903),(1303),(139530),(1333),(139551),(13950))x(a)ON RIGHT(LEFT(12335,n),1)=LEFT(a,1)ORDER BY n DESC SELECT LTRIM(i)+') '+REPLACE(IIF(L=4,'Fill ','Pour ')+RIGHT(a/100,L-3),9,j+' into ')+IIF(L=5,'tank',j)  +'
5L: '+LTRIM((A%100)/10)+', 3L: '+LTRIM(A%10)+', T: '+LTRIM(SUM(IIF(L=5,LEFT(a,1),0))OVER(ORDER BY i))FROM #t UNION SELECT 'Volume measured out in ' +LTRIM(MAX(i))+' turns'FROM #t

Input: 11

1) Fill 5L jug
5L: 5, 3L: 0, T: 0
2) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 5
3) Fill 5L jug
5L: 5, 3L: 0, T: 5
4) Pour from 5L jug into tank
5L: 0, 3L: 0, T: 10
5) Fill 3L jug
5L: 0, 3L: 3, T: 10
6) Pour from 3L jug into
5L jug 5L: 0, 3L: 3, T: 10
7) Fill 3L jug
5L: 3, 3L: 3, T: 10
8) Pour from 3L jug into 5L jug
5L: 5, 3L: 1, T: 10
9) Pour from 3L jug into tank
5L: 5, 3L: 0, T: 11
Volume measured out in 9 turns

Human-readable:

WITH n AS(
  SELECT 11 n
    UNION ALL
  SELECT n-IIF(n>4,5,3)
  FROM n
  WHERE n>2
)
SELECT n, a,LEN(a) L, i = IDENTITY(INT,1,1), 'L jug'j
INTO #t
FROM n
JOIN(VALUES
     (3303),(33900),
     (5550),(55900),
     (2550),(259323),(25903),
     (1303),(139530),(1333),(139551),(13950)
    )x(a)
ON RIGHT(LEFT(12335,n),1) = LEFT(a,1)
ORDER BY n DESC

 SELECT LTRIM(i)+') '
  + REPLACE(IIF(L=4,'Fill ','Pour ')
  + RIGHT(a/100,L-3),9,j+' into ')+IIF(L=5,'tank',j)
  +'
5L: ' + LTRIM((A%100)/10) + ', 3L: ' + LTRIM(A%10) + ', T: '
  + LTRIM(SUM(IIF(L=5,LEFT(a,1),0))OVER(ORDER BY i)) FROM #t
UNION ALL
 SELECT 'Volume measured out in ' +LTRIM(MAX(i))+' turns'FROM #t
 DROP TABLE #t

Answer 11

Python 3 (417 chars)

P=print
D=divmod
N=['3L jug','5L jug','tank',0]
M=999
R=[0,0,0,M]
F=[3,5,M,M]
def o(a,b):k=a==3;P(['Pour from %s into %s','Empty %s','Fill %s'][k*2+(b==3)]%[(N[a],N[b]),(N[b])][k]);d=min(R[a],F[b]-R[b]);R[a]-=d;R[b]+=d;P('5L:',R[1],'3L:',R[0],'T:',R[2]);N[3]+=1
k,r=D(int(input()),5)
for i in'0'*k:o(3,1);o(1,2)
for x in['','c1c12','d46','c2','d434d46'][r]:o(*D(int(x,16),4))
P('Volume measured out in',N[3],'turns')

Explained

Note that we have 4 objects, namely, the 3L jug, the 5L jug, the tank, and the foutain. The only operations we can do is to move water from object a to object b. This is what function o(a, b) do in my code, it move water and print it and keep counting.

Tricks

• N=['3L jug','5L jug','tank',0]. Here I need the last element to avoid IndexError. Also, it can be used as the global counting variable, without the expansive global keyword. For example, N[3] += 1

• Since 0 <= a < 4, 0 <= b < 4 in function o(a, b), we can encode (a, b) into a hex digit using (a << 2) | b, and decode it using divmod(x, 4). With this trick, all 5 solutions(reminder=0, 1, 2, 3, 4), can be encoded into array ['','c1c12','d46','c2','d434d46'], which is a little bit shorter than it's original form:

  A=[ (), ((3,0),(0,1),(3,0),(0,1),(0,2)), ((3,1),(1,0),(1,2)), ((3,0),(0,2)), ((3,1),(1,0),(0,3),(1,0),(3,1),(1,0),(1,2)) ]

Sample Output(n = 17)

17
Fill 5L jug
5L: 5 3L: 0 T: 0
Pour from 5L jug into tank
5L: 0 3L: 0 T: 5
Fill 5L jug
5L: 5 3L: 0 T: 5
Pour from 5L jug into tank
5L: 0 3L: 0 T: 10
Fill 5L jug
5L: 5 3L: 0 T: 10
Pour from 5L jug into tank
5L: 0 3L: 0 T: 15
Fill 5L jug
5L: 5 3L: 0 T: 15
Pour from 5L jug into 3L jug
5L: 2 3L: 3 T: 15
Pour from 5L jug into tank
5L: 0 3L: 3 T: 17
Volume measured out in 9 turns

Answer 12

COBOL (IBM Enterprise COBOL) 192 lines of 72 characters

This is a Proof of Concept for the Question, and the start of one for Golf-COBOL :-)

The question asks for the quickest. So, implement parallelism. Even one person can readily fill one 3L jug and one 5L jug at the same time.

Simply divide the input by eight, also leaving the remainder. Do some quick 5L/3L fills to the number of times eight fits exactly, then deal with the one through seven litres remaining.

The most interesting of the remainder is for four litres. Doing it as one litre plus three litres pushes a lot less water around, only 18 litres vs 23 for the other possibilities.

The Code (working)

   ID DIVISION
   PROGRAM-ID
   DATA DIVISION
   WORKING-STORAGE SECTION
   1.
   88 g1 VALUE ' '.
   2  PIC X
   88 H VALUE 'F'.
   88 I VALUE 'E'.
   88 J VALUE 'T'.
   2 PIC X
   88 K VALUE 'F'.
   88 L VALUE 'E'.
   88 M VALUE 'T'.
   1 R
   2 A1 PIC 999
   2 B PIC 99
   2 C PIC 9
   1 E
   2 e2 PIC X(120) VALUE "  ) Fill both jugs"
   2 e3 PIC X(120)
   88 O VALUE "5L: 0, 3L: 0, T: 000".
   2 e4 PIC X(120) VALUE "  ) Empty both jugs"
   2 e5 PIC X(120)
   2 e1 occurs 32 depending on p pic x(240)
   2 e6 pic x(99)
   1 F PIC 999 VALUE 0
   1 P PIC 99 VALUE 0
   1 P1 PIC 99
   PROCEDURE DIVISION
   ACCEPT A1
   DIVIDE A1 BY 8 GIVING B REMAINDER C
   set o to true
   move e3 to e5
   move 5 to e3(5:1)
   move 3 to e3(12:1)
   PERFORM D1 B TIMES
   EVALUATE C
   WHEN 1
   MOVE ZERO TO R
   SET K TO TRUE
   PERFORM N
   SET M TO TRUE
   PERFORM N
   SET K TO TRUE
   PERFORM N
   SET M TO TRUE
   PERFORM N
   SET L TO TRUE
   PERFORM N
   WHEN 2
   MOVE ZERO TO R
   SET H TO TRUE
   PERFORM N
   SET J TO TRUE
   PERFORM N
   SET I TO TRUE
   PERFORM N
   WHEN 3
   MOVE ZERO TO R
   SET K TO TRUE
   PERFORM N
   SET L TO TRUE
   PERFORM N
   WHEN 4
   MOVE ZERO TO R
   SET K TO TRUE
   PERFORM N
   SET M TO TRUE
   PERFORM N
   SET K TO TRUE
   PERFORM N
   SET M TO TRUE
   PERFORM N
   SET L TO TRUE
   PERFORM N
   SET K TO TRUE
   PERFORM N
   SET L TO TRUE
   PERFORM N
   WHEN 5
   MOVE ZERO TO R
   SET H TO TRUE
   PERFORM N
   SET I TO TRUE
   PERFORM N
   WHEN 6
   MOVE ZERO TO R
   SET K TO TRUE
   PERFORM N
   SET L TO TRUE
   PERFORM N
   SET K TO TRUE
   PERFORM N
   SET L TO TRUE
   PERFORM N
   WHEN 7
   MOVE ZERO TO R
   SET H TO TRUE
   PERFORM N
   SET I TO TRUE
   PERFORM N
   SET H TO TRUE
   PERFORM N
   SET J TO TRUE
   PERFORM N
   SET I TO TRUE
   PERFORM N
   END-EVALUATE
   string "Volume measured out in " delimited size P " turns"
   delimited size into e6
   if  e6(24:1) = 0
   move e6(25:) to e6 (24:)
   end-if
   move p to p1
   perform d2 p times
   DISPLAY E(481:)
   GOBACK
   D1
   ADD 1 TO P
   MOVE P TO E(1:2)
   move e2 to e1(p)
   move e3 to e1(p)(121:)
   ADD 1 TO P
   MOVE P TO E(241:2)
   ADD 8 TO F
   MOVE F TO E(378:3)
   move e4 to e1(p)
   move e5 to e1(p)(121:)
   MOVE F TO E(138:3)
   N
   ADD 1 TO P
   SET O TO TRUE
   EVALUATE TRUE
   WHEN K

   MOVE 3 TO B
   string p delimited size ") Fill 3L jug" delimited by size
   into e1(p)
   WHEN M
   COMPUTE C = C + B
   IF  C > 5
   COMPUTE B = C - 5
   MOVE 5 TO C
   ELSE
   MOVE 0 TO B
   END-IF
   string  P delimited size ") Pour from 3L jug into 5L jug"
   delimited size into e1(p)
   WHEN L
   ADD B TO F
   MOVE 0 TO B
   string  P delimited size ") Empty 3L jug into tank"
   delimited size into e1(p)
   END-EVALUATE
   EVALUATE TRUE
   WHEN H
   MOVE 5 TO C
   string  P delimited size ") Fill 5L jug"
   delimited size into e1(p)
   WHEN J
   COMPUTE B = C + B
   IF  B > 3
   COMPUTE C = B - 3
   MOVE 3 TO B
   ELSE
   MOVE 0 TO C
   END-IF
   string  P delimited size ") Pour from 5L jug into 3L jug"
   delimited size into e1(p)
   WHEN I
   ADD C TO F
   MOVE 0 TO C
   string  P delimited size ") Empty 5L jug into tank"
   delimited size into e1(p)
   END-EVALUATE
   string  "5L: " delimited size
       C delimited size ", 3L: " delimited size B(2:)
   ", T: " delimited size F delimited size
   into e1(p)(121:)
   SET g1 TO TRUE
   d2
   perform d3 2 times
   if  e1(p1)(1:1) = 0
   move e1(p1)(2:) to e1(p1)(1:120)
   end-if
   subtract 1 from p1
   d3
   if  e1(p1)(138:1) = 0
   move e1(p1)(139:) to e1(p1)(138:)
   end-if

This gets an absolute shed-load of diagnostic messages for code starting in the wrong place and shortage of required full-stops.

None of the diagnostics indicate any impact on the object code. So, despite it being a busted RC=8 I know the object will be OK, so linked it and ran it.

Here are the outputs for one to eight litres. After that, all results can be intuited. 17 and 100 are included as examples of the parallelism.

There is still much that can be done to squeeze the program down in characters, the correct output was the important thing first. Counting the characters when they are on fixed-length lines is another thing entirely.

Sample output:

1) Fill 3L jug                 
5L: 0, 3L: 3, T: 0             
2) Pour from 3L jug into 5L jug
5L: 3, 3L: 0, T: 0             
3) Fill 3L jug                 
5L: 3, 3L: 3, T: 0             
4) Pour from 3L jug into 5L jug
5L: 5, 3L: 1, T: 0             
5) Empty 3L jug into tank      
5L: 5, 3L: 0, T: 1             
Volume measured out in 5 turns 

1) Fill 5L jug                 
5L: 5, 3L: 0, T: 0             
2) Pour from 5L jug into 3L jug
5L: 2, 3L: 3, T: 0             
3) Empty 5L jug into tank      
5L: 0, 3L: 3, T: 2             
Volume measured out in 3 turns

1) Fill 3L jug                 
5L: 0, 3L: 3, T: 0             
2) Empty 3L jug into tank      
5L: 0, 3L: 0, T: 3             
Volume measured out in 2 turns 

1) Fill 3L jug                 
5L: 0, 3L: 3, T: 0             
2) Pour from 3L jug into 5L jug
5L: 3, 3L: 0, T: 0             
3) Fill 3L jug                 
5L: 3, 3L: 3, T: 0             
4) Pour from 3L jug into 5L jug
5L: 5, 3L: 1, T: 0             
5) Empty 3L jug into tank      
5L: 5, 3L: 0, T: 1             
6) Fill 3L jug                 
5L: 5, 3L: 3, T: 1             
7) Empty 3L jug into tank      
5L: 5, 3L: 0, T: 4             
Volume measured out in 7 turns 

1) Fill 5L jug                
5L: 5, 3L: 0, T: 0            
2) Empty 5L jug into tank     
5L: 0, 3L: 0, T: 5            
Volume measured out in 2 turns

1) Fill 3L jug                 
5L: 0, 3L: 3, T: 0             
2) Empty 3L jug into tank      
5L: 0, 3L: 0, T: 3             
3) Fill 3L jug                 
5L: 0, 3L: 3, T: 3             
4) Empty 3L jug into tank      
5L: 0, 3L: 0, T: 6             
Volume measured out in 4 turns 

1) Fill 5L jug                  
5L: 5, 3L: 0, T: 0              
2) Empty 5L jug into tank       
5L: 0, 3L: 0, T: 5              
3) Fill 5L jug                  
5L: 5, 3L: 0, T: 5              
4) Pour from 5L jug into 3L jug 
5L: 2, 3L: 3, T: 5              
5) Empty 5L jug into tank       
5L: 0, 3L: 3, T: 7              
Volume measured out in 5 turns 



1) Fill both jugs               
5L: 5, 3L: 3, T: 0              
2) Empty both jugs              
5L: 0, 3L: 0, T: 8              
Volume measured out in 2 turns  

1) Fill both jugs               
5L: 5, 3L: 3, T: 0              
2) Empty both jugs              
5L: 0, 3L: 0, T: 8              
3) Fill both jugs               
5L: 5, 3L: 3, T: 8              
4) Empty both jugs              
5L: 0, 3L: 0, T: 16             
5) Fill 3L jug                  
5L: 0, 3L: 3, T: 16             
6) Pour from 3L jug into 5L jug 
5L: 3, 3L: 0, T: 16             
7) Fill 3L jug                  
5L: 3, 3L: 3, T: 16             
8) Pour from 3L jug into 5L jug 
5L: 5, 3L: 1, T: 16             
9) Empty 3L jug into tank       
5L: 5, 3L: 0, T: 17             
Volume measured out in 9 turns  



1) Fill both jugs  
5L: 5, 3L: 3, T: 0 
2) Empty both jugs 
5L: 0, 3L: 0, T: 8 
3) Fill both jugs  
5L: 5, 3L: 3, T: 8 
4) Empty both jugs 
5L: 0, 3L: 0, T: 16
5) Fill both jugs  
5L: 5, 3L: 3, T: 16
6) Empty both jugs 
5L: 0, 3L: 0, T: 24
7) Fill both jugs  
5L: 5, 3L: 3, T: 24
8) Empty both jugs 
5L: 0, 3L: 0, T: 32
9) Fill both jugs  
5L: 5, 3L: 3, T: 32
10) Empty both jugs
5L: 0, 3L: 0, T: 40
11) Fill both jugs 
5L: 5, 3L: 3, T: 40
12) Empty both jugs
5L: 0, 3L: 0, T: 48
13) Fill both jugs 
5L: 5, 3L: 3, T: 48
14) Empty both jugs
5L: 0, 3L: 0, T: 56
15) Fill both jugs 
5L: 5, 3L: 3, T: 56
16) Empty both jugs
5L: 0, 3L: 0, T: 64
17) Fill both jugs 
5L: 5, 3L: 3, T: 64
18) Empty both jugs
5L: 0, 3L: 0, T: 72
19) Fill both jugs               
5L: 5, 3L: 3, T: 72              
20) Empty both jugs              
5L: 0, 3L: 0, T: 80              
21) Fill both jugs               
5L: 5, 3L: 3, T: 80              
22) Empty both jugs              
5L: 0, 3L: 0, T: 88              
23) Fill both jugs               
5L: 5, 3L: 3, T: 88              
24) Empty both jugs              
5L: 0, 3L: 0, T: 96              
25) Fill 3L jug                  
5L: 0, 3L: 3, T: 96              
26) Pour from 3L jug into 5L jug 
5L: 3, 3L: 0, T: 96              
27) Fill 3L jug                  
5L: 3, 3L: 3, T: 96              
28) Pour from 3L jug into 5L jug 
5L: 5, 3L: 1, T: 96              
29) Empty 3L jug into tank       
5L: 5, 3L: 0, T: 97              
30) Fill 3L jug                  
5L: 5, 3L: 3, T: 97              
31) Empty 3L jug into tank       
5L: 5, 3L: 0, T: 100             
Volume measured out in 31 turns